Dice
Problem Description
You have a dice with m faces, each face contains a distinct number. We assume when we tossing the dice, each face will occur randomly and uniformly. Now you have T query to answer, each query has one of the following form:
0 m n: ask for the expected number of tosses until the last n times results are all same.
1 m n: ask for the expected number of tosses until the last n consecutive results are pairwise different.
Input
The first line contains a number T.(1≤T≤100) The next T line each line contains a query as we mentioned above. (1≤m,n≤106) For second kind query, we guarantee n≤m. And in order to avoid potential precision issue, we guarantee the result for our query
will not exceeding 109 in this problem.
Output
For each query, output the corresponding result. The answer will be considered correct if the absolute or relative error doesn‘t exceed 10-6.
Sample Input
6 0 6 1 0 6 3 0 6 5 1 6 2 1 6 4 1 6 6 10 1 4534 25 1 1232 24 1 3213 15 1 4343 24 1 4343 9 1 65467 123 1 43434 100 1 34344 9 1 10001 15 1 1000000 2000
Sample Output
1.000000000 43.000000000 1555.000000000 2.200000000 7.600000000 83.200000000 25.586315824 26.015990037 15.176341160 24.541045769 9.027721917 127.908330426 103.975455253 9.003495515 15.056204472 4731.706620396
Source
2013 Multi-University Training Contest 5
题目大意：
n边形的骰子，问你出现连续相同（不同）n次需要掷的次数的数学期望。
解题思路：
利用递归方式的DP的思想推公式
（1）若询问为0，则：
dp[i] 记录的是已经连续i个相同，到n个不同需要的次数的数学期望
dp[0]= 1+dp[1]
dp[1]= 1+( 1/m*dp[2]+(m-1)/m*dp[1])=1+(dp[2]+(1-m)*dp[1])/m;
dp[2]= 1+(dp[3]+(1-m)*dp[1])/m;
....................
dp[n]= 0
推出：
dp[i] = 1 + ( (m-1)*dp[1] + dp[i+1] ) / m
dp[i+1] = 1 + ( (m-1)*dp[1] + dp[i+2] ) / m
因此，m*(dp[i+1]-dp[i])=(dp[i+2]-dp[i+1])
我们发现是等比数列
dp[0]-dp[1]=1;
dp[1]-dp[2]=m;
..........
dp[n-1]-dp[n]=m^(n-1)
累加，得：dp[0]-dp[n]=1+m+m^2+..........m^(n-1)=(1-m^n)/(1-m)
所以：dp[0]=(1-m^n)/(1-m);
（2）若询问为1，则：
dp[0] = 1 + dp[1]
dp[1] = 1 + (dp[1] + (m-1) dp[2]) / m
dp[2] = 1 + (dp[1] + dp[2] + (m-2) dp[3]) / m
dp[i] = 1 + (dp[1] + dp[2] + ... dp[i] + (m-i)*dp[i+1]) / m
dp[i+1]= 1 + (dp[1] + dp[2] + ... dp[i] + dp[i+1] + (m-i-1)*dp[i+1]) / m
...
dp[n] = 0;
选出 dp[i] 和 dp[i+1] 这两行相减 得
dp[i] - dp[i+1] = (m-i-1)/m * (dp[i+1] - dp[i+2]);
因此 dp[i+1] - dp[i+2] = m/(m-i-1)*(dp[i]-dp[i+1]);
所以：
dp[0]-dp[1]=1;
dp[1]-dp[2]=1*m/(m-1);
dp[2]-dp[3]=1*m/(m-1)*m/(m-2);
..........
dp[n-1]-dp[n]=1*m/(m-1)*m/(m-2)*.......*m/(m-n+1);
累加得到答案
解题代码：
#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
inline double solve(){
int op,m,n;
scanf("%d%d%d",&op,&m,&n);
double ans=0;
if(op==0){
for(int i=0;i<=n-1;i++){
ans+=pow(1.0*m,i);
}
}else{
double tmp=1.0;
for(int i=1;i<=n;i++){
ans+=tmp;
tmp*=m*1.0/(m-i);
}
}
return ans;
}
int main(){
int t;
while(scanf("%d",&t)!=EOF){
while(t-- >0){
printf( "%.9lf\n",solve() );
}
}
return 0;
}
HDU 4652 Dice （概率DP）,布布扣,bubuko.com